Parameterized complexity of constraint satisfaction problems D - - PowerPoint PPT Presentation

Parameterized complexity of constraint satisfaction problems

D´ aniel Marx Budapest University of Technology and Economics

dmarx@cs.bme.hu

Presented at the University of Newcastle, Australia March 15, 2004

Parameterized complexity of constraint satisfaction problems – p.1/21

Constraint satisfaction problems

Let R be a set Boolean of relations. An R-formula is a conjuction of relations in R:

R1(x1, x4, x5) ∧ R2(x2, x1) ∧ R1(x3, x3, x3) ∧ R3(x5, x1, x4, x1) R-SAT

Given: an R-formula ϕ Find: a variable assignment satisfying ϕ

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Constraint satisfaction problems

Let R be a set Boolean of relations. An R-formula is a conjuction of relations in R:

R1(x1, x4, x5) ∧ R2(x2, x1) ∧ R1(x3, x3, x3) ∧ R3(x5, x1, x4, x1) R-SAT

Given: an R-formula ϕ Find: a variable assignment satisfying ϕ

R = {a = b} ⇒ R-SAT = 2-coloring of a graph R = {a ∨ b, a ∨ ¯ b, ¯ a ∨ ¯ b} ⇒ R-SAT = 2SAT R = {a ∨ b ∨ c, a ∨ b ∨ ¯ c, a ∨ ¯ b ∨ ¯ c, ¯ a ∨ ¯ b ∨ ¯ c} ⇒ R-SAT = 3SAT

Question: R-SAT is polynomial time solvable for which R? It is NP-complete for which R?

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Schaefer’s Dichotomy Theorem (1978)

For every R, the R-SAT problem is polynomial time solvable if one of the following holds, and NP-complete otherwise: Every relation is satisfied by the all 0 assignment Every relation is satisfied by the all 1 assignment Every relation can be expressed by a 2SAT formula Every relation can be expressed by a Horn formula Every relation can be expressed by an anti-Horn formula Every relation is an affine subspace over GF (2)

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Schaefer’s Dichotomy Theorem (1978)

For every R, the R-SAT problem is polynomial time solvable if one of the following holds, and NP-complete otherwise: Every relation is satisfied by the all 0 assignment Every relation is satisfied by the all 1 assignment Every relation can be expressed by a 2SAT formula Every relation can be expressed by a Horn formula Every relation can be expressed by an anti-Horn formula Every relation is an affine subspace over GF (2) Why is it surprising?

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Ladner’s Theorem (1975)

If P = NP, then there is a language L ∈ NP \ P that is not NP-complete.

P=NP P P NP NP

NP−complete NP−complete NP−intermediate

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Other dichotomy results

Approximability of MAX-SAT, MIN-UNSAT [Khanna et al., 2001] Approximability of MAX-ONES, MIN-ONES [Khanna et al., 2001] Generalization to 3 valued variables [Bulatov, 2002] Inverse satisfiability [Kavvadias and Sideri, 1999] etc.

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Other dichotomy results

Approximability of MAX-SAT, MIN-UNSAT [Khanna et al., 2001] Approximability of MAX-ONES, MIN-ONES [Khanna et al., 2001] Generalization to 3 valued variables [Bulatov, 2002] Inverse satisfiability [Kavvadias and Sideri, 1999] etc. Our contribution: parameterized analogue of Schaefer’s dichotomy theorem.

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Parameterized version

Parameterized R-SAT Input: a R-formula ϕ, an integer k Parameter: k Question: Does ϕ has a satisfying assignment of weight exactly k? For which R is there an f(k) · nc algorithm for R-SAT? Main theorem: For every constraint family R, the parameter R-SAT problem is either fixed-parameter tractable or W[1]-complete. (+ simple characterization of FPT cases)

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Technical notes

Are constants allowed in the formula? E.g., R(x1, 0, 1) ∧ R(1, x2, x3) Can a variable appear multiple times in a constraint? E.g., R(x1, x1, x2) ∧ R(x3, x3, x3) Constraints that are not satisfied by the all 0 assignment can be handled easily (bounded search tree).

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Weak separability

Definition: R is weakly separable if

• 1. the union of two disjoint satisfying assignments is also satisfying, and
• 2. if a satisfying assignment contains a smaller satisfying assignment, then their

difference is also satisfying. Example of 1:

R(1, 1, 1, 1, 0, 0, 0, 0, 0) = 1 R(0, 0, 0, 0, 1, 1, 0, 0, 0) = 1 ⇓ R(1, 1, 1, 1, 1, 1, 0, 0, 0) = 1

Example of 2:

R(1, 1, 1, 1, 1, 1, 0, 0) = 1 R(0, 0, 1, 1, 1, 1, 0, 0) = 1 ⇓ R(1, 1, 0, 0, 0, 0, 0, 0) = 1

Main theorem: R-SAT is FPT if and only if every constraint is weakly separable, and W[1]-complete otherwise.

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Weak separability: examples

The constraint EVEN is weakly separable: Property 1:

R(

even

• 1, 1, 1, 1, 0, 0, 0, 0, 0) = 1

R(0, 0, 0, 0, 1, 1

• even

, 0, 0, 0) = 1 ⇓ R(1, 1, 1, 1, 1, 1

• even

, 0, 0, 0) = 1

Property 2:

R(

even

• 1, 1, 1, 1, 1, 1, 0, 0) = 1

R(0, 0, 1, 1, 1, 1

• even

, 0, 0) = 1 ⇓ R(1, 1

• even

, 0, 0, 0, 0, 0, 0) = 1

More generally: every affine constraint is weakly separable.

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Weak separability: examples (cont.)

The following constraint is trivially weakly separable:

R(0, 0, 0, 0, 0) = 1 R(1, 1, 1, 0, 0) = 1 R(0, 1, 1, 1, 0) = 1 R(0, 0, 1, 1, 1) = 1 R(x1, x2, x3, x4, x5) = 0 otherwise.

Reason: Property 1 and 2 vacously hold, no disjoint sets, no subsets. More generally: if the non-zero satisfying assignments are intersecting and form a clutter, then it is weakly separable. Example: R(x1, . . . , xn) = 1 if and only if 0 or exactly t out of n variables are 1 (t > n/2)

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Parameterized vs. classical

The easy and hard cases are different in the classacial and the parameterized version: Constraint Classical Parameterized

x ∨ y

in P FPT (Vertex Cover)

¯ x ∨ ¯ y

in P W[1]-complete (Max. Independent Set) affine in P FPT 2-in-3 NP-complete FPT

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Bounded number of occurrences

Primal graph: Vertices are the variables, two variables are connected if they appear in some clause together.

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Bounded number of occurrences

Primal graph: Vertices are the variables, two variables are connected if they appear in some clause together. Every satisfying assignment is composed of connected satisfying assignments. Lemma: There are at most (rd)k2 · n connected satisfying assignment of size at most k. (r is the maximum arity, d is the maximum no. of occurences) Algorithm: Use color coding to put together the connected assignments to obtain a size k assignment.

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The sunflower lemma

Definition: Sets S1, S2, . . . , Sk form a sunflower if the sets

Si \ (S1 ∩ S2 ∩ · · · ∩ Sk) are disjoint.

petals center Lemma (Erd˝

• s and Rado, 1960): If the size of a set system is greater than

(p − 1)ℓ · ℓ! and it contains only sets of size at most ℓ, then the system contains a

sunflower with k petals.

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Sunflower of clauses

Definition: A sunflower is a set of k clauses such that for every i either the same variable appears at position i in every clause,

• r every clause “owns” its ith variable.

R(x1, x2, x3, x4, x5, x6) R(x1, x2, x3, x7, x8, x9) R(x1, x2, x3, x10, x11, x12) R(x1, x2, x3, x13, x14, x15)

Lemma: If a variable occurs more than cR(k) times in an R-formula, then there is a sunflower of clauses with more than k petals in the formula.

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Plucking the sunflower

For weakly separable constraints, the formula can be reduced if there is a sunflower with k + 1 petals. Example:

k + 1         

EVEN(x1, x2, x3, x4, x5, x6) EVEN(x1, x2, x3, x7, x8, x9) EVEN(x1, x2, x3, x10, x11, x12) EVEN(x1, x2, x3, x13, x14, x15)

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Plucking the sunflower

For weakly separable constraints, the formula can be reduced if there is a sunflower with k + 1 petals. Example:

k + 1         

EVEN(x1, x2, x3, x4, x5, x6) EVEN(x1, x2, x3, x7, x8, x9) EVEN(x1, x2, x3, 0, 0, 0) EVEN(x1, x2, x3, x13, x14, x15)

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Plucking the sunflower

For weakly separable constraints, the formula can be reduced if there is a sunflower with k + 1 petals. Example:

k + 1         

EVEN(x1, x2, x3, x4, x5, x6) EVEN(x1, x2, x3, x7, x8, x9) EVEN(x1, x2, x3, 0, 0, 0) EVEN(x1, x2, x3, x13, x14, x15)

⇓

EVEN(x1, x2, x3)

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Plucking the sunflower

For weakly separable constraints, the formula can be reduced if there is a sunflower with k + 1 petals. Example:

k + 1         

EVEN(x1, x2, x3, x4, x5, x6) EVEN(x1, x2, x3, x7, x8, x9) EVEN(x1, x2, x3, 0, 0, 0) EVEN(x1, x2, x3, x13, x14, x15)

⇓

EVEN(x1, x2, x3) EVEN(x4, x5, x6) EVEN(x7, x8, x9) EVEN(x10, x11, x12) EVEN(x13, x14, x15)

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The algorithm

If there is a variable that occurs more than cR(k) times: Find a sunflower with k + 1 petals Pluck the sunflower ⇒ shorter formula If every variable occurs at most cR(k) times: Apply the bounded occurence algorithm Running time: 2kr+2·22O(r)

· n log n, where r is the maximum arity in the

constraint family R.

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Hardness results: case 1

Definition: R is weakly separable if

• 1. the union of two disjoint satisfying assignments is also satisfying, and
• 2. if a satisfying assignment contains a smaller satisfying assignment, then their

difference is also satisfying. If property 1 is violated:

R(0, 0, 0, 0, 0, 0, 0, 0) = 1 R(1, 1, 1, 0, 0, 0, 0, 0) = 1 R(0, 0, 0, 1, 1, 0, 0, 0) = 1 R(1, 1, 1, 1, 1, 0, 0, 0) = 0

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Hardness results: case 1

Definition: R is weakly separable if

• 1. the union of two disjoint satisfying assignments is also satisfying, and
• 2. if a satisfying assignment contains a smaller satisfying assignment, then their

difference is also satisfying. If property 1 is violated:

R(0, 0, 0, 0, 0, 0, 0, 0) = 1 R(1, 1, 1, 0, 0, 0, 0, 0) = 1 R(0, 0, 0, 1, 1, 0, 0, 0) = 1 R(1, 1, 1, 1, 1, 0, 0, 0) = 0 ⇓ R(x, x, x, y, y, 0, 0, 0) = 1 ⇐ ⇒ ¯ x ∨ ¯ y

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Hardness results: case 1

Definition: R is weakly separable if

• 1. the union of two disjoint satisfying assignments is also satisfying, and
• 2. if a satisfying assignment contains a smaller satisfying assignment, then their

difference is also satisfying. If property 1 is violated:

R(0, 0, 0, 0, 0, 0, 0, 0) = 1 R(1, 1, 1, 0, 0, 0, 0, 0) = 1 R(0, 0, 0, 1, 1, 0, 0, 0) = 1 R(1, 1, 1, 1, 1, 0, 0, 0) = 0 ⇓

Maximum Independent Set

R(x, x, x, y, y, 0, 0, 0) = 1 ⇐ ⇒ ¯ x ∨ ¯ y ⇒ can be expressed!

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Hardness results: case 2

Definition: R is weakly separable if

• 1. the union of two disjoint satisfying assignments is also satisfying, and
• 2. if a satisfying assignment contains a smaller satisfying assignment, then their

difference is also satisfying. If property 2 is violated:

R(0, 0, 0, 0, 0, 0, 0, 0) = 1 R(1, 1, 1, 1, 1, 0, 0, 0) = 1 R(0, 0, 0, 1, 1, 0, 0, 0) = 1 R(1, 1, 1, 0, 0, 0, 0, 0) = 0

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Hardness results: case 2

Definition: R is weakly separable if

• 1. the union of two disjoint satisfying assignments is also satisfying, and
• 2. if a satisfying assignment contains a smaller satisfying assignment, then their

difference is also satisfying. If property 2 is violated:

R(0, 0, 0, 0, 0, 0, 0, 0) = 1 R(1, 1, 1, 1, 1, 0, 0, 0) = 1 R(0, 0, 0, 1, 1, 0, 0, 0) = 1 R(1, 1, 1, 0, 0, 0, 0, 0) = 0 ⇓ R(x, x, x, y, y, 0, 0, 0) = 1 ⇐ ⇒ x → y

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Hardness results: case 2

Definition: R is weakly separable if

• 1. the union of two disjoint satisfying assignments is also satisfying, and
• 2. if a satisfying assignment contains a smaller satisfying assignment, then their

difference is also satisfying. If property 2 is violated:

R(0, 0, 0, 0, 0, 0, 0, 0) = 1

Lemma: The problem is

R(1, 1, 1, 1, 1, 0, 0, 0) = 1

W[1]-complete for the

R(0, 0, 0, 1, 1, 0, 0, 0) = 1

constraint →.

R(1, 1, 1, 0, 0, 0, 0, 0) = 0 ⇓ R(x, x, x, y, y, 0, 0, 0) = 1 ⇐ ⇒ x → y

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Planar formulae

If the primal graph of the formula is planar, then the layering method of Baker can be used.

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Planar formulae

If the primal graph of the formula is planar, then the layering method of Baker can be used. Set to 0 the variables in every (k + 1)th layer. There are k + 1 ways of doing this. One of them will not hurt the solution. Example with k = 3:

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Planar formulae

If the primal graph of the formula is planar, then the layering method of Baker can be used. Set to 0 the variables in every (k + 1)th layer. There are k + 1 ways of doing this. One of them will not hurt the solution. Example with k = 3:

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Planar formulae (cont.)

If we delete every (k + 1)th layer, then the remaining formula has only k layers: Lemma (Bodlaender): The treewidth of a k-layered graph is at most 3k − 1. If the primal graph has bounded treewidth, then the problem can be solved in linear time using standard techniques.

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Planar formulae (cont.)

If we delete every (k + 1)th layer, then the remaining formula has only k layers: Lemma (Bodlaender): The treewidth of a k-layered graph is at most 3k − 1. If the primal graph has bounded treewidth, then the problem can be solved in linear time using standard techniques. Incidence graph: bipartite graph, vertices are the clauses and the variable, edge means “appears in”. Theorem: Linear time alg. if the incidence graph of the formula is planar.

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Summary

Parameterized version of R-SAT FPT or W[1]-complete depending on weak separabilty Bounded occurences: color coding using connected solutions Reduction using the sunflower lemma Linear time solvable for planar and bounded treewidth formulae

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Summary

Parameterized version of R-SAT FPT or W[1]-complete depending on weak separabilty Bounded occurences: color coding using connected solutions Reduction using the sunflower lemma Linear time solvable for planar and bounded treewidth formulae

Thank you for your attention! Questions?

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